Problem: $\sum\limits_{n=1}^{\infty}\dfrac{1}{n+3}$ When applying the integral test, we get a limit that determines whether the series converges or diverges. What is this limit? Choose 1 answer: Choose 1 answer: (Choice A) A $\lim_{b\to\infty}\ln\left(\dfrac{1}{b+3}\right)$ (Choice B) B $\lim_{b\to\infty}\dfrac{b+4}{b+3}$ (Choice C) C $\lim_{b\to\infty}\dfrac{1}{b+3}$ (Choice D) D $\lim_{b\to\infty}\ln\left(\dfrac{b+3}{4}\right)$
Solution: $\dfrac{1}{x+3}$ satisfies the conditions for the integral test. This means that $\sum\limits_{n=1}^{\infty}\dfrac{1}{n+3}$ converges/diverges together with $\int_1^{\infty}\dfrac{1}{x+3}\,dx$. $\int_1^{\infty}\dfrac{1}{x+3}\,dx=\lim_{b\to\infty}\ln\left(\dfrac{b+3}{4}\right)$ In conclusion, the limit that determines whether the series converges or diverges is $\lim_{b\to\infty}\ln\left(\dfrac{b+3}{4}\right)$.